Question

If \( n \) is an integer and \( Z = \cos \theta + i \sin \theta, \theta \neq (2n + 1)\frac{\pi}{2}, \) then: \[ \frac{1 + Z^{2n}}{1 - Z^{2n}} = ? \]

• A. \( i \tan n\theta \)
• B. \( i \cot n\theta \)
• C. \( -i \tan n\theta \)
• D. \( -i \cot n\theta \)

Hint:

When working with complex exponentials, apply De Moivre's theorem to express the powers of complex numbers in terms of sines and cosines. Use the conjugate to simplify the fraction.

Answer 1

The Correct Option is B

We are given \( Z = \cos \theta + i \sin \theta \), which is the polar form of a complex number, and we are asked to evaluate: \[ \frac{1 + Z^{2n}}{1 - Z^{2n}}. \] Using De Moivre's theorem, we know: \[ Z^{2n} = \cos(2n\theta) + i \sin(2n\theta). \] Step 1: Substituting for \( Z^{2n} \) Substituting into the expression: \[ \frac{1 + \cos(2n\theta) + i \sin(2n\theta)}{1 - \cos(2n\theta) - i \sin(2n\theta)}. \] Step 2: Simplifying the expression Multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{(1 + \cos(2n\theta) + i \sin(2n\theta)) (1 - \cos(2n\theta) + i \sin(2n\theta))}{(1 - \cos(2n\theta) - i \sin(2n\theta)) (1 - \cos(2n\theta) + i \sin(2n\theta))}. \] Simplifying the denominator: \[ (1 - \cos(2n\theta))^2 + \sin^2(2n\theta) = 2(1 - \cos(2n\theta)) = 4\sin^2(n\theta). \] The numerator simplifies to: \[ i \cot n\theta. \] Thus, the value of the given expression is \( i \cot n\theta \).